Answer
$$s\left( t \right) = \frac{4}{3}{t^{3/2}} + 1$$
Work Step by Step
$$\eqalign{
& v\left( t \right) = 2\sqrt t ;{\text{ }}s\left( 0 \right) = 1 \cr
& {\text{Calculating the position function}} \cr
& s\left( t \right) = \int {v\left( t \right)} dt \cr
& s\left( t \right) = \int {\left( {2\sqrt t } \right)} dt \cr
& s\left( t \right) = 2\left( {\frac{{{t^{3/2}}}}{{3/2}}} \right) + C \cr
& s\left( t \right) = \frac{4}{3}{t^{3/2}} + C \cr
& {\text{Calculating the function for the initial position }}s\left( 0 \right) = 1 \cr
& 1 = \frac{4}{3}{\left( 0 \right)^{3/2}} + C \cr
& 1 = C \cr
& {\text{Thenerefore,}} \cr
& s\left( t \right) = \frac{4}{3}{t^{3/2}} + 1 \cr
& {\text{Graphing the velocity and position functions}} \cr} $$
